Frustums of Cones and Pyramids

If the top of a cone or a pyramid is removed, eliminating the figure’s vertex, the result is a frustum.

Geometry Lesson 103

A frustum of a cone is a part of a cone with two parallel circular bases.

A frustum of a pyramid is a part of a pyramid with two parallel bases.

Geometry Lesson 103

Volume of a Frustum - The following formula is used to find the volume of a frustum, regardless of whether it is part of a cone or a pyramid. The variables 𝐵1 and 𝐵2 are the areas of the two bases and h is the height of the frustum.

𝑉 =1

3ℎ 𝐵1 + 𝐵1 + 𝐵2 + 𝐵2

Geometry Lesson 103

Find the volume of the frustum of the pyramid shown. SOLUTION Find the area of each base of the frustum.

𝐵1 = 𝑏ℎ 𝐵2 = 𝑏ℎ 𝐵1 = 8 10 𝐵2 = 6 7.5

𝐵1 = 80 𝐵2 = 45 Now, apply the formula for volume of a frustum.

𝑉 =1

3ℎ 𝐵1 + 𝐵1 + 𝐵2 + 𝐵2

𝑉 =1

310 80 + 80 + 45 + 45

𝑉 = 6162

3

The volume of this frustum is 6162

3 cubic meters.

Geometry Lesson 103

a. Find the volume of the frustum of the cone to the nearest hundredth of a cubic inch. SOLUTION Notice the two triangles highlighted in the diagram. Both are right triangles that share an angle. Therefore, they are similar. Write a proportion.

𝑟

15=

14

30

𝑟 = 7

Now find the area of the bases. 𝐵1 = 𝜋𝑟2 𝐵2 = 𝜋𝑟2

𝐵1 = 𝜋72 𝐵2 = 𝜋142 𝐵1 ≈ 153.94 𝐵2 ≈ 615.75

Finally, apply the formula for volume of a frustum.

𝑉 =1

3ℎ 𝐵1 + 𝐵1 + 𝐵2 + 𝐵2

𝑉 ≈1

315 153.94 + 153.94 + 615.75 + 615.75

𝑉 ≈ 5387.84

The volume is approximately 5387.84 cubic inches.

Geometry Lesson 103

b. Find the areas of the frustum’s bases. SOLUTION Since one angle of the cone’s cross section is given, and the cross section is a right triangle, the tangent function can be used to find x, the radius of the lower base.

tan 70° =15

𝑥

𝑥 ≈ 5.46 Now find the area of the bases.

𝐵1 = 𝜋𝑟2 𝐵2 = 𝜋𝑟2 𝐵1 = 𝜋32 𝐵2 ≈ 𝜋5.462 𝐵1 ≈ 28.27 𝐵2 ≈ 93.66

So the areas of the frustum’s bases are approximately 28.27 square centimeters and 93.66 square centimeters.

Geometry Lesson 103

A grain silo is shaped like a cone, as shown in the diagram. If the height of the grain in the silo is 40 feet, what volume of grain is in the silo, to the nearest cubic foot?

SOLUTION

As in Example 2a, similar triangles will have to be used to find the radius of the top of the frustum made by the grain.

The diagram illustrates the similar triangles.

Write a proportion. 𝑟

40=

20

80

𝑟 = 10

Next, find the area of the frustum’s bases. 𝐵1 = 𝜋𝑟2 𝐵2 = 𝜋𝑟2

𝐵1 = 𝜋102 𝐵2 ≈ 𝜋202 𝐵1 ≈ 314.16 𝐵2 ≈ 1256.64

Finally, apply the formula for volume of a frustum.

𝑉 =1

3ℎ 𝐵1 + 𝐵1 + 𝐵2 + 𝐵2

𝑉 ≈1

340 314.16 + 314.16 + 1256.64 + 1256.64

𝑉 ≈ 29322

The volume of grain in the silo is approximately 29,322 cubic feet.

Geometry Lesson 103

a. Find the volume of this frustum of a pyramid. Round your answer to the nearest cubic inch.

Geometry Lesson 103

b. Find the volume of this frustum of a cone to the nearest hundredth.

Geometry Lesson 103

c. The Great Pyramid of Giza is the largest of ancient Egypt’s pyramids. It is a square pyramid that stands 147 meters tall. The diagram depicts what the Great Pyramid might have looked like during construction. Given the dimensions in the diagram, what is the volume of the pyramid at this point in its construction?

Geometry Lesson 103

Page 670

Lesson Practice (Ask Mr. Heintz)

Page 671

Practice 1-30 (Do the starred ones first)

Geometry Lesson 103